CODEWITHSUNDEEP
phpmysqli
WebbieWorks
WebbieWorks

Reputation: 163

mysqli insert data is blank and adding double

I am trying to learn mysqli so I convert my site over. And I was having a hard time on the update part so I decided to take baby steps and go a little at a time.

Well I am trying to switch over the add-item page which uses the INSERT. The issue is, there is nothing being put into the database and it putting 2 blank rows in. (Like 2 items are being added)

I have been wearing this board out with questions and I search for the answers and do trial & error before asking, but I just cannot seem to grasp this mysqli. It works great with mysql, but with it being depreciated, I wanted to switch it over.

post2.php 

     <?php

        $db = new mysqli("localhost","admin","pass","database"); 

        if(!$db) {
    die('sorry we are having some problbems');
}

if(isset($_POST['submit'])) {
    $stmt = $db->prepare("INSERT INTO new_equip(`id`,`itemname`, `manufacture`, `model`,`serial`) VALUES(?,?,?,?,?)");
    $stmt->bind_param("issss", $_POST['id'], $_POST['itemname'], $_POST['manufacture'], $_POST['model'], $_POST['serial']);

    if(!$stmt->execute()) {
    // Do your error stuff

  die('Error: ' . mysqli_error($db));
 }

}
echo $sql;
var_dump($_POST);
mysqli_close($db);
?>

add-item.php 

<form method="post" action="post2.php" enctype="multipart/form-data" class="form-horizontal" accept-charset="UTF-8">
<div class="form-group">
    <label class="col-md-3">Item Name</label>
    <input type="text" name="itemname" value="" class="form-control" />
</div> <!-- /.form-group -->
<div class="form-group">
    <label class="col-md-3">Manufacture</label>
    <input type="text" name="manufacture" value="" class="form-control" />
</div> <!-- /.form-group -->
<div class="form-group">
    <label class="col-md-3">Model</label>
    <input type="text" name="model" value="" class="form-control" />
</div> <!-- /.form-group -->
<div class="form-group">
    <label class="col-md-3">Serial</label>
    <input type="text" name="serial" value="" class="form-control" />
</div> <!-- /.form-group -->
<div class="col-md-8 col-md-push-3">
    <input type="submit" name="Submit" class="btn btn-primary">
    &nbsp;
    <button type="reset" class="btn btn-default">Cancel</button>
</div> <!-- /.col -->

Upvotes: 0

Views: 954

Answers (1)

AlexL
AlexL

Reputation: 1707

That's because you run the query 2 times.

One time here:

$result = $db->query($sql);

Second time here:

if (!mysqli_query($db,$sql))

You should replace if (!mysqli_query($db,$sql)) with if (!$result) or just remove $result = $db->query($sql);

Prepared statement version:

<?php

    $db = new mysqli("localhost","admin","pass","database"); 

    if(!$db) {
    die('sorry we are having some problbems');
}

if(isset($_POST['submit'])) {
    $stmt = $db->prepare("INSERT INTO new_equip(`id`,`itemname`, `manufacture`, `model`,`serial`) VALUES(?,?,?,?,?)");
    $stmt->bind_param("issss", $_POST['id'], $_POST['itemname'], $_POST['manufacture'], $_POST['model'], $_POST['serial']);

    if(!$stmt->execute()) {
        // Do your error stuff

        die('Error: ' . mysqli_error($db));
    }

}

echo $sql;
var_dump($_POST);
mysqli_close($db);
?>

Upvotes: 2

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